A beam of protons with speed 4 × 10⁵ ms –1 enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. The pitch of the resulting helical path of protons is close to :
(Mass of the proton = 1.67 × 10 –27 kg, charge
of the proton = 1.69 × 10 –19 C)

Solution

Physics sir · 3 steps

Split the velocity: the part along B carries the proton forward, the part across B makes it circle.

1. Split the velocity. The component along the field, v cos 60° = 4 × 10⁵ × 0.5 = 2 × 10⁵ m s⁻¹, feels no force and stays constant. The component across the field, v sin 60°, makes the proton go round in a circle.
2. Time for one turn. The period T = (2π m)/qB does not depend on speed. Here qB = 1.69 × 10⁻¹⁹ × 0.3 = 5.07 × 10⁻²⁰, so T = (2π × 1.67 × 10⁻²⁷)/(5.07 × 10⁻²⁰) ≈ 2.07 × 10⁻⁷ s.

   p = (v cos θ) T = 2 × 10⁵ × 2.07 × 10⁻⁷ ≈ 4.1 × 10⁻² m

3. So. The pitch is about 4.1 cm, which is closest to 4 cm, option (d).

The idea to keep. Pitch is the distance moved along B in one full turn: p = v cos θ × (2π m)/qB. Use the parallel component for pitch and the perpendicular component for radius.

Answer from the NTA final answer key. Solution by PrepWiser.